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November 17th, 2015, 11:35 PM  #1 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7  Solve the equation completely.
Solve the equation completely. $$4^{(2x+3/2)}65(4^x)+8=0$$ How do I tackle this? Last edited by Chikis; November 17th, 2015 at 11:55 PM. 
November 18th, 2015, 12:42 AM  #2 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 
4^(2x + 3/2)  65(4^x) + 8 = 0 ((4^2x) * 4^(3/2))  65(4^x) + 8 = 0 [[(4^x)^2] * 8]  65(4^x) + 8 = 0 let 4^x = a... Then the expression becomes... 8a^2 65a + 8 = 0 8a^2  64a  a + 8 = 0 8a(a  8 ) 1(a  8 ) = 0 (8a  1)(a  8 ) = 0 so, a = 1/8 or a = 8 so, a = 4^x = 1/8 or 4^x = 8 4^x = 2^2x = 2^3 or 2^2x = 2^3 so... 2x = 3 or 2x = 3 x = 3/2 or x = 3/2 
November 18th, 2015, 04:32 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra  
November 18th, 2015, 05:10 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980  
November 18th, 2015, 08:48 AM  #5  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
* $\displaystyle \ \ \ \ $4^2x = (4^2)x $\displaystyle \ \ \ \ $You don't mean that. ** $\displaystyle \ \ $2^2x = (2^2)x $\displaystyle \ \ \ \ $You don't mean that.  
November 30th, 2015, 07:55 PM  #6 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7 
Thanks from Chikis.


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